php 只有变量应该通过引用传递

Only variables should be passed by reference

提问by Frank Nwoko

// Other variables
$MAX_FILENAME_LENGTH = 260;
$file_name = $_FILES[$upload_name]['name'];
//echo "testing-".$file_name."<br>";
//$file_name = strtolower($file_name);
$file_extension = end(explode('.', $file_name)); //ERROR ON THIS LINE
$uploadErrors = array(
    0=>'There is no error, the file uploaded with success',
    1=>'The uploaded file exceeds the upload max filesize allowed.',
    2=>'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form',
    3=>'The uploaded file was only partially uploaded',
    4=>'No file was uploaded',
    6=>'Missing a temporary folder'
);

Any ideas? After 2 days still stuck.

有任何想法吗？2天后仍然卡住。

回答by Oswald

Assign the result of explodeto a variable and pass that variable to end:

将 的结果分配explode给变量并将该变量传递给end：

$tmp = explode('.', $file_name);
$file_extension = end($tmp);

The problem is, that endrequires a reference, because it modifies the internal representation of the array (i.e. it makes the current element pointerpoint to the last element).

问题是，这end需要引用，因为它修改了数组的内部表示（即它使当前元素指针指向最后一个元素）。

The result of explode('.', $file_name)cannot be turned into a reference. This is a restriction in the PHP language, that probably exists for simplicity reasons.

的结果explode('.', $file_name)不能转化为参考。这是 PHP 语言中的一个限制，可能出于简单原因而存在。

回答by Sinan Eldem

Php 7 compatible proper usage:

PHP 7 兼容正确用法：

$fileName      = 'long.file.name.jpg';
$tmp           = explode('.', $fileName);
$fileExtension = end($tmp);

echo $fileExtension;
// jpg

回答by ryeguy

Everyone else has already given you the reason you're getting an error, but here's the best way to do what you want to do: $file_extension = pathinfo($file_name, PATHINFO_EXTENSION);

其他人都已经向您提供了出现错误的原因，但这是执行您想做的事情的最佳方法： $file_extension = pathinfo($file_name, PATHINFO_EXTENSION);

回答by Floern

save the array from explode() to a variable, and then call end() on this variable:

将explode()中的数组保存到一个变量中，然后在这个变量上调用end()：

$tmp = explode('.', $file_name);
$file_extension = end($tmp);

btw: I use this code to get the file extension:

顺便说一句：我使用此代码来获取文件扩展名：

$ext = substr( strrchr($file_name, '.'), 1);

where strrchrextracts the string after the last .and substrcuts off the .

wherestrrchr提取最后一个之后的字符串.并substr切断.

回答by Will Vousden

Try this:

尝试这个：

$parts = explode('.', $file_name);
$file_extension = end($parts);

The reason is that the argument for endis passed by reference, since endmodifies the array by advancing its internal pointer to the final element. If you're not passing a variable in, there's nothing for a reference to point to.

原因是 for 的参数end是通过引用传递的，因为end通过将其内部指针前进到最后一个元素来修改数组。如果你没有传入一个变量，就没有什么可以指向的引用。

See endin the PHP manual for more info.

有关end更多信息，请参阅PHP 手册。

回答by wizard

PHP complains because end()expects a reference to something that it wants to change (which can be a variable only). You however pass the result of explode()directly to end()without saving it to a variable first. At the moment when explode()returns your value, it exists only in memory and no variable points to it. You cannot create a reference to something (or to something unknown in the memory), that does not exists.

PHP 抱怨是因为end()期望引用它想要更改的内容（只能是变量）。但是，您可以explode()直接将结果传递给 toend()而无需先将其保存到变量中。在explode()返回值的那一刻，它只存在于内存中，没有变量指向它。你不能创建对不存在的东西（或内存中未知的东西）的引用。

Or in other words: PHP does not know, if the value you give him is the direct value or just a pointer to the value (a pointer is also a variable (integer), which stores the offset of the memory, where the actual value resides). So PHP expects here a pointer (reference) always.

或者换句话说：PHP不知道，你给他的值是直接值还是只是一个指向该值的指针（指针也是一个变量（整数），它存放的是内存的偏移量，实际值在哪里居住）。所以 PHP 总是希望这里有一个指针（引用）。

But since this is still just a notice (not even deprecated) in PHP 7, you can savely ignore notices and use the ignore-operator instead of completely deactivating error reporting for notices:

但是由于这在 PHP 7 中仍然只是一个通知（甚至没有被弃用），您可以节省地忽略通知并使用 ignore-operator 而不是完全停用通知的错误报告：

$file_extension = @end(explode('.', $file_name));

回答by Tgr

end(...[explode('.', $file_name)])has worked since PHP 5.6. This is documented in the RFCalthough not in PHP docs themselves.

end(...[explode('.', $file_name)])从 PHP 5.6 开始工作。这在RFC 中有记录，但不在 PHP 文档中。

回答by jon_darkstar

Just as you can't index the array immediately, you can't call end on it either. Assign it to a variable first, then call end.

正如你不能立即索引数组一样，你也不能调用 end 。先赋值给一个变量，然后调用end。

$basenameAndExtension = explode('.', $file_name);
$ext = end($basenameAndExtension);

回答by NVRM

Since it raise a flag for over 10 years, but works just fine and return the expected value, a little stfu operatoris the goodiest bad practice you are all looking for:

由于它已升旗超过 10 年，但工作正常并返回预期值，因此一点stfu 运算符是您正在寻找的最糟糕的做法：

$file_extension = @end(explode('.', $file_name));

回答by evenvi

PHP offical Manual : end()

PHP 官方手册：end()

Parameters

参数

array

The array. This array is passed by reference because it is modified by the function. This means you must pass it a real variable and not a function returning an array because only actual variables may be passed by reference.

阵列。这个数组是通过引用传递的，因为它被函数修改了。这意味着您必须向它传递一个真正的变量，而不是一个返回数组的函数，因为只有实际变量才能通过引用传递。